🗓️ Mock for Week 5 and 6

❓ Question 1

(Using information from the image with questions 1, 2, and 3)

Question: Let the random variables $X$ and $Y$ have the joint density function: $f_{XY}(x, y) = \begin{cases} 1 & \text{for } 0 \le x < 1, 0 \le y < 1 \ 0 & \text{otherwise} \end{cases}$

Calculate $P\left(0 < X < \frac{1}{2}, \frac{1}{4} < Y < \frac{1}{2}\right)$. Enter the answers correct to three decimal places.

💡 Concept: Probability as Area for Uniform Distributions

This joint PDF $f_{XY}(x, y)$ describes a uniform distribution over the unit square (from $x=0$ to $x=1$ and $y=0$ to $y=1$).

For a continuous joint PDF, you find probability by calculating the volume under the PDF curve over the region of interest.

Because our PDF has a constant height of 1 inside the unit square, this integral simplifies. The probability is just the area of the region of interest, as long as that region is inside the unit square.

Solution

1. Identify the Region: The problem asks for the probability of the rectangular region $A$ defined by $0 < X < \frac{1}{2}$ and $\frac{1}{4} < Y < \frac{1}{2}$.
2. Check Support: This entire region is inside the support (the $1 \times 1$ unit square).
3. Calculate the Area: The region is a rectangle.
   • $\text{Width} = (\text{upper } x) - (\text{lower } x) = \frac{1}{2} - 0 = \frac{1}{2}$
   • $\text{Height} = (\text{upper } y) - (\text{lower } y) = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$
4. Area = $\text{Width} \times \text{Height} = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$
5. Find Probability: Since the PDF height is 1 in this region, the probability is equal to the area.
   • $P = \text{Area} = \frac{1}{8}$
6. Format Answer: To convert to a decimal, $1 \div 8 = 0.125$.

Answer: 0.125

❓ Question 2

(Using information from the image with questions 1, 2, and 3)

Question: Find $P\left(0 < X < \frac{1}{2}\right)$. Enter the answer correct to one decimal place.

💡 Concept: Marginal Probability from a Joint PDF

This question asks for a marginal probability—the probability of $X$ being in a range, regardless of the value of $Y$. To find this, we consider the entire possible range of $Y$ within the support.

The region of interest is $0 < X < \frac{1}{2}$ and (implicitly) $0 \le Y < 1$ (the full range of $Y$ from the PDF’s definition).

Solution

1. Identify the Region: The region is defined by $0 < X < \frac{1}{2}$ and $0 \le Y < 1$.
2. Calculate the Area: This is again a simple rectangle.
   • $\text{Width} = (\text{upper } x) - (\text{lower } x) = \frac{1}{2} - 0 = \frac{1}{2}$
   • $\text{Height} = (\text{upper } y) - (\text{lower } y) = 1 - 0 = 1$
3. Area = $\text{Width} \times \text{Height} = \frac{1}{2} \times 1 = \frac{1}{2}$
4. Find Probability: $P = \text{Area} = \frac{1}{2}$
5. Format Answer: $\frac{1}{2} = 0.5$.

Answer: 0.5

❓ Question 3

(Using information from the image with questions 1, 2, and 3)

Question: Find $P(X < 2Y)$. Enter the answer correct to two decimal places.

💡 Concept: Probability of a Non-Rectangular Region

Here, the region of interest is not a simple rectangle. It’s the area within the unit square that also satisfies the inequality $X < 2Y$. We can rewrite the inequality as $Y > \frac{X}{2}$. We need to find the area of the portion of the unit square that is above the line $Y = \frac{X}{2}$.

Solution

1. Sketch the Region:

   • Draw the $1 \times 1$ unit square (our support).
   • Draw the line $Y = \frac{X}{2}$. This line starts at $(0, 0)$ and goes to $(1, \frac{1}{2})$.
   • The region $Y > \frac{X}{2}$ is the area above this line.
   • We want the area of this region inside the square.

2. Calculate the Area: It’s easier to calculate the area of the small triangle below the line and subtract it from the total area of the square.

   • $\text{Area of unit square} = 1 \times 1 = 1$.
   • The unwanted area (below the line $Y = \frac{X}{2}$) is a triangle with vertices at $(0, 0)$, $(1, 0)$, and $(1, \frac{1}{2})$.
   • $\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height}$
   • $\text{Area of triangle} = \frac{1}{2} \times 1 \times \frac{1}{2} = \frac{1}{4}$

3. Find the Desired Area:

   • $\text{Area}(Y > \frac{X}{2}) = \text{Area}(\text{Square}) - \text{Area}(\text{Triangle})$
   • $\text{Area}(Y > \frac{X}{2}) = 1 - \frac{1}{4} = \frac{3}{4}$

4. Find Probability: $P(X < 2Y) = \text{Area} = \frac{3}{4}$

5. Format Answer: $\frac{3}{4} = 0.75$.

Answer: 0.75

❓ Question 4

(Using information from the image with questions 4 and 5)

Question: Let $X$ be a continuous random variable with PDF $f_X(x) = \begin{cases} 2/3, & 0 < x < 1 \ 1/3, & 2 < x < 3 \ 0, & \text{otherwise} \end{cases}$ Which among the following represent the cumulative distribution function (CDF) of $X$?

💡 Concept: Deriving the CDF from the PDF

The Cumulative Distribution Function (CDF), $F(x)$, is the probability $P(X \le x)$. It is the integral of the Probability Density Function (PDF), $f(t)$, from $-\infty$ to $x$.

Key properties:

• The CDF $F(x)$ is the “running total” of probability.
• The value of the PDF $f(x)$ is the slope of the CDF $F(x)$.
• Where $f(x) = 0$ (like between $x=1$ and $x=2$), the CDF $F(x)$ will be flat (slope = 0).
• Where $f(x)$ is a positive constant (like $2/3$), the CDF $F(x)$ will be a straight line with that constant as its slope.

This means the correct CDF graph must be piecewise linear, not a step function. This immediately rules out the bottom graph. Let’s build the correct CDF (the top graph) step-by-step.

Solution

1. For $x \le 0$: The PDF $f(x)$ is 0. The total accumulated probability is 0. $F(x) = 0$. The graph starts at $(-\infty, 0)$ and goes to $(0, 0)$.

2. For $0 < x < 1$: The PDF $f(x)$ is $2/3$. The CDF increases with a slope of $2/3$. $F(x) = F(0) + \int_{0}^{x} \frac{2}{3} ,dt = 0 + \frac{2}{3}x = \frac{2}{3}x$. At the end of this interval, $F(1) = \frac{2}{3}(1) = \frac{2}{3}$. The graph is a straight line from $(0, 0)$ to $(1, 2/3)$.

3. For $1 \le x < 2$: The PDF $f(x)$ is 0. The slope of the CDF is 0, so it’s a flat horizontal line. It stays at the last probability it reached. $F(x) = F(1) = \frac{2}{3}$. The graph is a flat line from $(1, 2/3)$ to $(2, 2/3)$.

4. For $2 \le x < 3$: The PDF $f(x)$ is $1/3$. The CDF now increases again, with a slope of $1/3$. $F(x) = F(2) + \int_{2}^{x} \frac{1}{3} ,dt = \frac{2}{3} + \left[ \frac{1}{3}t \right]_2^x = \frac{2}{3} + (\frac{1}{3}x - \frac{2}{3}) = \frac{1}{3}x$. At the end of this interval, $F(3) = \frac{1}{3}(3) = 1$. The graph is a straight line from $(2, 2/3)$ to $(3, 1)$.

5. For $x \ge 3$: The PDF $f(x)$ is 0. The slope is 0. The CDF has accumulated all possible probability (a total of 1) and stays flat. $F(x) = 1$. The graph is a flat line at $y=1$ for all $x \ge 3$.

Answer: The top graph is the correct representation of the CDF.

❓ Question 5

(Using information from the image with questions 4 and 5)

Question: Find the value of $P(X \le 2.5)$. Enter the answer correct to two decimal places.

💡 Concept: Using the CDF to Find Probability

The CDF $F(x)$ is defined as $P(X \le x)$. Therefore, finding $P(X \le 2.5)$ is as simple as finding the value of the CDF $F(x)$ at $x = 2.5$.

We can use the piecewise function we derived in Question 4.

Solution

1. Identify the Interval: The value $x = 2.5$ falls into the interval $2 \le x < 3$.
2. Find the CDF Formula: In the previous question, we found that for $2 \le x < 3$, the CDF is: $F(x) = \frac{1}{3}x$ (Alternatively, $F(x) = F(2) + (\text{probability from 2 to x}) = \frac{2}{3} + \int_{2}^{x} \frac{1}{3} ,dt = \frac{2}{3} + \frac{1}{3}(x-2)$)
3. Evaluate at $x = 2.5$: Let’s use the second, more intuitive formula:
   • $F(2.5) = F(2) + (\text{probability from 2 to 2.5})$
   • $F(2.5) = \frac{2}{3} + \int_{2}^{2.5} \frac{1}{3} ,dt$
   • $F(2.5) = \frac{2}{3} + \left[ \frac{1}{3}t \right]_2^{2.5}$
   • $F(2.5) = \frac{2}{3} + \left( \frac{1}{3}(2.5) - \frac{1}{3}(2) \right)$
   • $F(2.5) = \frac{2}{3} + \left( \frac{2.5}{3} - \frac{2}{3} \right)$
   • $F(2.5) = \frac{2}{3} + \frac{0.5}{3} = \frac{2.5}{3}$
   • $F(2.5) = \frac{5/2}{3} = \frac{5}{6}$
4. Convert to Decimal: $5 \div 6 = 0.8333…$
5. Format Answer: Rounding to two decimal places, we get $0.83$.

Answer: 0.83

❓ Question 6

(Using information from the image with question 6)

Question: Suppose random variables $X$ and $Y$ are uniformly distributed over the region $D$, where $D = {(x, y): [0, 2] \times [-2, 0] \cup [-1, 0] \times [0, 1]}$. Choose the correct options from the following:

• $f_{XY}(x, y) = \begin{cases} 4, & 0 < x < 2, -2 < y < 0 \ 1, & -1 < x < 0, 0 < y < 1 \ 0, & \text{otherwise} \end{cases}$
• $f_{XY}(x, y) = \begin{cases} \frac{1}{5}, & x, y \in D \ 0, & \text{otherwise} \end{cases}$
• $f_{Y|X=1}(-1) = 0.5$
• $f_{Y|X=1}(-1) = 0$
• $f_{Y|X=1}(-1) = 0.625$

💡 Concept 1: Joint PDF for a Uniform Distribution

For a distribution to be uniform over a region $D$, its PDF $f_{XY}(x, y)$ must be a constant value $c$ for all points inside $D$. For the total probability (volume) to equal 1, this constant $c$ must be $1 / (\text{Total Area of } D)$.

💡 Concept 2: Conditional PDF

The conditional PDF $f_{Y|X}(y|x)$ is the PDF of $Y$ given that $X$ is fixed at a specific value $x$. For a uniform distribution, this is found by:

1. Taking a vertical “slice” of the region $D$ at the given $x$.
2. The conditional distribution $f_{Y|X}(y|x)$ is a new uniform distribution, but only over that vertical slice.
3. Its height will be $1 / (\text{Length of the slice})$.

Solution

Part 1: Find the Joint PDF $f_{XY}(x, y)$

1. Calculate the Total Area of $D$: The region $D$ is made of two rectangles.
   • $\text{Rectangle 1}$ (bottom-right): $0 \le x \le 2$ and $-2 \le y \le 0$. $\text{Area}_1 = \text{width} \times \text{height} = (2 - 0) \times (0 - (-2)) = 2 \times 2 = 4$.
   • $\text{Rectangle 2}$ (top-left): $-1 \le x \le 0$ and $0 \le y \le 1$. $\text{Area}_2 = \text{width} \times \text{height} = (0 - (-1)) \times (1 - 0) = 1 \times 1 = 1$.
2. Total Area = $\text{Area}_1 + \text{Area}_2 = 4 + 1 = 5$.
3. Find PDF Height: Since the distribution is uniform, the PDF’s height must be $1 / (\text{Total Area})$. $f_{XY}(x, y) = \frac{1}{5}$ for all $(x, y)$ in $D$.
4. Evaluate Options:
   • The first option is incorrect. It suggests different densities in different parts of $D$.
   • The second option, $f_{XY}(x, y) = \begin{cases} \frac{1}{5}, & x, y \in D \ 0, & \text{otherwise} \end{cases}$, is correct.

Part 2: Find the Conditional PDF $f_{Y|X=1}(-1)$

1. Condition: We are given $X = 1$.
2. Find the “Slice”: Look at the region $D$ and draw a vertical line at $X=1$. This line only passes through the bottom-right rectangle.
3. Find the Length of the Slice: At $X=1$, the possible $Y$ values are from $-2$ to $0$.
   • Length of slice = (top $y$) - (bottom $y$) = $0 - (-2) = 2$.
4. Determine Conditional PDF: The conditional distribution of $Y$ given $X=1$ is uniform over the interval $[-2, 0]$. Its height is $1 / (\text{Length of slice})$.
   • $f_{Y|X=1}(y) = \frac{1}{2}$, for $-2 \le y \le 0$.
5. Evaluate at $y = -1$: We need to find $f_{Y|X=1}(-1)$.
   • Since $y = -1$ is inside the interval $[-2, 0]$, the value of the PDF is $\frac{1}{2}$.
   • $\frac{1}{2} = 0.5$.
6. Evaluate Options:
   • $f_{Y|X=1}(-1) = 0.5$ is correct.
   • The other two conditional options are incorrect.

Answer: The two correct options are:

• $f_{XY}(x, y) = \begin{cases} \frac{1}{5}, & x, y \in D \ 0, & \text{otherwise} \end{cases}$
• $f_{Y|X=1}(-1) = 0.5$

❓ Question 7

(Using information from the image with questions 7 and 8)

Question: 30% of the total players in IPL 2022 are uncapped… and 70% are capped… Suppose the runs scored by the capped players is Normal(60, 25) and the runs scored by the uncapped players is Normal(55, 36).

Find the distribution of runs of a randomly chosen player.

💡 Concept: Law of Total Probability (Mixture Distributions)

The distribution of a randomly chosen player is a mixture distribution. Its PDF is a weighted average of the PDFs of the two subgroups (capped and uncapped).

The general formula is: $f_{\text{Total}}(y) = P(\text{Capped}) \cdot f_{\text{Capped}}(y) + P(\text{Uncapped}) \cdot f_{\text{Uncapped}}(y)$

We also need the formula for a Normal distribution PDF: $f(y) = \frac{1}{\sigma\sqrt{2\pi}} \exp\left(-\frac{(y - \mu)^2}{2\sigma^2}\right)$

Solution

1. Identify Components:

   • Capped Players:
     • $P(\text{Capped}) = 0.7 = \frac{7}{10}$
     • Distribution: Normal($\mu_C = 60, \sigma_C^2 = 25$) $\implies \sigma_C = 5$
     • PDF: $f_{\text{Capped}}(y) = \frac{1}{5\sqrt{2\pi}} \exp\left(-\frac{(y - 60)^2}{2 \cdot 25}\right) = \frac{1}{5\sqrt{2\pi}} \exp\left(-\frac{(y - 60)^2}{50}\right)$
   • Uncapped Players:
     • $P(\text{Uncapped}) = 0.3 = \frac{3}{10}$
     • Distribution: Normal($\mu_U = 55, \sigma_U^2 = 36$) $\implies \sigma_U = 6$
     • PDF: $f_{\text{Uncapped}}(y) = \frac{1}{6\sqrt{2\pi}} \exp\left(-\frac{(y - 55)^2}{2 \cdot 36}\right) = \frac{1}{6\sqrt{2\pi}} \exp\left(-\frac{(y - 55)^2}{72}\right)$

2. Build the Mixture PDF: $f_{\text{Total}}(y) = (0.7) \cdot f_{\text{Capped}}(y) + (0.3) \cdot f_{\text{Uncapped}}(y)$ $f_{\text{Total}}(y) = \left(\frac{7}{10}\right) \left[ \frac{1}{5\sqrt{2\pi}} \exp\left(-\frac{(y - 60)^2}{50}\right) \right] + \left(\frac{3}{10}\right) \left[ \frac{1}{6\sqrt{2\pi}} \exp\left(-\frac{(y - 55)^2}{72}\right) \right]$

3. Simplify the Coefficients:

   • First term’s coefficient: $\frac{7}{10} \times \frac{1}{5} = \frac{7}{50}$
   • Second term’s coefficient: $\frac{3}{10} \times \frac{1}{6} = \frac{3}{60} = \frac{1}{20}$

4. Final Equation: $f_{\text{Total}}(y) = \frac{7}{50\sqrt{2\pi}} \exp\left(-\frac{(y - 60)^2}{50}\right) + \frac{1}{20\sqrt{2\pi}} \exp\left(-\frac{(y - 55)^2}{72}\right)$

Answer: This matches the second option.

❓ Question 8

(Using information from the image with questions 7 and 8)

Question: If a randomly selected player scored 60 runs, what is the probability that the selected candidate is a capped player? Enter the answer correct to two decimal places.

💡 Concept: Bayes’ Theorem (for Continuous Variables)

This is a conditional probability question. We want to find $P(\text{Capped} | Y = 60)$, where $Y$ is the random variable for runs.

Bayes’ Theorem states: $P(A|B) = \frac{P(B|A)P(A)}{P(B)}$

For our problem, this becomes: $P(\text{Capped} | Y=60) = \frac{f(Y=60 | \text{Capped}) \cdot P(\text{Capped})}{f_{\text{Total}}(Y=60)}$

• $f(Y=60 | \text{Capped})$ is just the Capped PDF $f_{\text{Capped}}(60)$.
• $P(\text{Capped})$ is the prior probability, $0.7$.
• $f_{\text{Total}}(Y=60)$ is the total PDF from Question 7, evaluated at $y=60$.

Solution

1. Numerator: $f_{\text{Capped}}(60) \cdot P(\text{Capped})$

   • $f_{\text{Capped}}(y) = \frac{1}{5\sqrt{2\pi}} \exp\left(-\frac{(y - 60)^2}{50}\right)$
   • $f_{\text{Capped}}(60) = \frac{1}{5\sqrt{2\pi}} \exp\left(-\frac{(60 - 60)^2}{50}\right) = \frac{1}{5\sqrt{2\pi}} \exp(0) = \frac{1}{5\sqrt{2\pi}}$
   • $\text{Numerator} = \left( \frac{1}{5\sqrt{2\pi}} \right) \cdot (0.7) = \frac{0.7}{5\sqrt{2\pi}}$

2. Denominator: $f_{\text{Total}}(60)$ This is $f_{\text{Capped}}(60) \cdot P(\text{Capped}) + f_{\text{Uncapped}}(60) \cdot P(\text{Uncapped})$

   • We already have the first part: $\frac{0.7}{5\sqrt{2\pi}}$
   • Let’s find the second part: $f_{\text{Uncapped}}(60) \cdot P(\text{Uncapped})$
     • $f_{\text{Uncapped}}(y) = \frac{1}{6\sqrt{2\pi}} \exp\left(-\frac{(y - 55)^2}{72}\right)$
     • $f_{\text{Uncapped}}(60) = \frac{1}{6\sqrt{2\pi}} \exp\left(-\frac{(60 - 55)^2}{72}\right) = \frac{1}{6\sqrt{2\pi}} \exp\left(-\frac{25}{72}\right)$
     • $\text{Second Part} = \left( \frac{1}{6\sqrt{2\pi}} \exp\left(-\frac{25}{72}\right) \right) \cdot (0.3) = \frac{0.3}{6\sqrt{2\pi}} \exp\left(-\frac{25}{72}\right)$
   • $\text{Denominator} = \frac{0.7}{5\sqrt{2\pi}} + \frac{0.3}{6\sqrt{2\pi}} \exp\left(-\frac{25}{72}\right)$

3. Calculate the Probability: $P(\text{Capped} | Y=60) = \frac{\frac{0.7}{5\sqrt{2\pi}}}{\frac{0.7}{5\sqrt{2\pi}} + \frac{0.3}{6\sqrt{2\pi}} \exp\left(-\frac{25}{72}\right)}$

   We can cancel $\frac{1}{\sqrt{2\pi}}$ from the numerator and denominator: $P(\text{Capped} | Y=60) = \frac{\frac{0.7}{5}}{\frac{0.7}{5} + \frac{0.3}{6} \exp\left(-\frac{25}{72}\right)}$

4. Plug in Values:

   • $\frac{0.7}{5} = 0.14$
   • $\frac{0.3}{6} = 0.05$
   • $\exp(-\frac{25}{72}) \approx \exp(-0.34722) \approx 0.70664$
   • $P(\text{Capped} | Y=60) = \frac{0.14}{0.14 + (0.05 \cdot 0.70664)}$
   • $P(\text{Capped} | Y=60) = \frac{0.14}{0.14 + 0.035332}$
   • $P(\text{Capped} | Y=60) = \frac{0.14}{0.175332} \approx 0.79848$

5. Format Answer: Rounding to two decimal places, we get 0.80.

Answer: 0.80

❓ Question 9

(Using information from the image with questions 9, 10, and 11)

Question: Suppose the time to failure… is exponentially distributed… Let $X$ and $Y$ denote the time to failure of Devices $A$ and $B$, respectively. The joint pdf of $X$ and $Y$ is given by $f_{XY}(x, y) = \begin{cases} k e^{-(4x+5y)} & \text{if } x > 0, y > 0 \ 0 & \text{otherwise} \end{cases}$

Find the value of $k$.

💡 Concept: Normalizing a PDF

For any probability density function (PDF), the total integral (or “volume”) under the function over its entire support (the region where it’s non-zero) must be equal to 1. We can use this fact to solve for the normalizing constant $k$.

Solution

1. Set up the Integral: $\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f_{XY}(x, y) ,dx ,dy = 1$ $\int_{0}^{\infty} \int_{0}^{\infty} k e^{-(4x+5y)} ,dx ,dy = 1$

2. Separate the Variables: We can rewrite $e^{-(4x+5y)}$ as $e^{-4x} \cdot e^{-5y}$. $k \int_{0}^{\infty} \int_{0}^{\infty} e^{-4x} e^{-5y} ,dx ,dy = 1$ $k \left( \int_{0}^{\infty} e^{-4x} ,dx \right) \left( \int_{0}^{\infty} e^{-5y} ,dy \right) = 1$

3. Solve the Integrals:

   • $\int_{0}^{\infty} e^{-4x} ,dx = \left[ -\frac{1}{4} e^{-4x} \right]_0^\infty = (0) - (-\frac{1}{4} e^0) = \frac{1}{4}$
   • $\int_{0}^{\infty} e^{-5y} ,dy = \left[ -\frac{1}{5} e^{-5y} \right]_0^\infty = (0) - (-\frac{1}{5} e^0) = \frac{1}{5}$

4. Solve for $k$: $k \left( \frac{1}{4} \right) \left( \frac{1}{5} \right) = 1$ $k \left( \frac{1}{20} \right) = 1$ $k = 20$

Answer: 20

❓ Question 10

(Using information from the image with questions 9, 10, and 11)

Question: Find the value of $\alpha + \beta$. (Given $X \sim \text{Exp}(\alpha)$ and $Y \sim \text{Exp}(\beta)$)

💡 Concept: Independence and Marginal Distributions

The PDF of an exponential distribution with parameter $\lambda$ is $f(z) = \lambda e^{-\lambda z}$.

• $f_X(x) = \alpha e^{-\alpha x}$
• $f_Y(y) = \beta e^{-\beta y}$

If $X$ and $Y$ are independent, their joint PDF is the product of their marginal PDFs: $f_{XY}(x,y) = f_X(x) \cdot f_Y(y)$

Solution

1. Check for Independence: Our joint PDF is $f_{XY}(x,y) = 20 e^{-4x} e^{-5y}$. This function can be separated into a product of a function of only $x$ and a function of only $y$. This means $X$ and $Y$ are independent.

2. Find the Marginals: $f_{XY}(x,y) = (C_1 e^{-4x}) \cdot (C_2 e^{-5y})$, where $C_1 \cdot C_2 = 20$.

   • For $f_X(x) = C_1 e^{-4x}$ to be a valid PDF, it must integrate to 1. $\int_{0}^{\infty} C_1 e^{-4x} ,dx = 1 \implies C_1 \cdot (\frac{1}{4}) = 1 \implies C_1 = 4$.
   • For $f_Y(y) = C_2 e^{-5y}$ to be a valid PDF, it must integrate to 1. $\int_{0}^{\infty} C_2 e^{-5y} ,dy = 1 \implies C_2 \cdot (\frac{1}{5}) = 1 \implies C_2 = 5$. (Check: $C_1 \cdot C_2 = 4 \cdot 5 = 20$, which matches our $k$!)

3. Identify Parameters:

   • The marginal for $X$ is $f_X(x) = 4 e^{-4x}$.
   • Comparing this to the definition $f_X(x) = \alpha e^{-\alpha x}$, we see that $\alpha = 4$.
   • The marginal for $Y$ is $f_Y(y) = 5 e^{-5y}$.
   • Comparing this to the definition $f_Y(y) = \beta e^{-\beta y}$, we see that $\beta = 5$.

4. Calculate the Sum: $\alpha + \beta = 4 + 5 = 9$.

Answer: 9

❓ Question 11

(Using information from the image with questions 9, 10, and 11)

Question: Find the probability that Device $B$ will last longer when compared to Device $A$. Enter the answer correct to two decimal places.

💡 Concept: Probability of $Y > X$

We need to find $P(Y > X)$. This can be solved in two ways:

1. Integration: Integrate the joint PDF $f_{XY}(x,y)$ over the region where $y > x$ (and $x>0, y>0$). $P(Y > X) = \int_{0}^{\infty} \int_{x}^{\infty} 20 e^{-4x-5y} ,dy ,dx$

2. Shortcut for Independent Exponentials: For two independent exponential random variables $X \sim \text{Exp}(\alpha)$ and $Y \sim \text{Exp}(\beta)$, there is a standard result: $P(X < Y) = \frac{\alpha}{\alpha + \beta}$ (The probability that the one with parameter $\alpha$ “wins” is $\frac{\alpha}{\alpha + \beta}$)

The question asks for $P(Y > X)$, which is the same as $P(X < Y)$.

Solution

Using the shortcut (Method 2) is much faster.

1. Identify Parameters (from Q10):

   • $\alpha = 4$
   • $\beta = 5$

2. Apply the Formula: $P(Y > X) = P(X < Y) = \frac{\alpha}{\alpha + \beta}$ $P(Y > X) = \frac{4}{4 + 5} = \frac{4}{9}$

3. Format Answer: $4 \div 9 = 0.4444…$ Rounding to two decimal places, we get 0.44.

(Self-Check with Integration - Method 1):

• $\text{Inner integral (w.r.t } y\text{): } \int_{x}^{\infty} 20 e^{-4x} e^{-5y} ,dy = 20 e^{-4x} \left[ -\frac{1}{5} e^{-5y} \right]_{y=x}^{y=\infty}$ $= 20 e^{-4x} \left( (0) - (-\frac{1}{5} e^{-5x}) \right) = 4 e^{-4x} e^{-5x} = 4 e^{-9x}$
• $\text{Outer integral (w.r.t } x\text{): } \int_{0}^{\infty} 4 e^{-9x} ,dx = 4 \left[ -\frac{1}{9} e^{-9x} \right]_0^\infty$ $= 4 \left( (0) - (-\frac{1}{9} e^0) \right) = 4 \left( \frac{1}{9} \right) = \frac{4}{9}$.
• The result is confirmed.

Answer: 0.44

❓ Question 12

(Using information from the image with questions 12 and 13)

Question: Let $X$ be a continuous uniform random variable on $[0, 1]$ and $Y = \frac{1}{X}$. Find the probability density function of $Y$.

💡 Concept: Transformation of Variables (CDF Method)

We can find the PDF of $Y$ by first finding its Cumulative Distribution Function (CDF), $F_Y(y)$, and then differentiating it. $f_Y(y) = \frac{d}{dy} F_Y(y)$.

1. Find the support (range) of $Y$:

   • $X$ is in $[0, 1]$.
   • When $X$ is close to 0 (e.g., $X=0.01$), $Y = 1/0.01 = 100$ (large). As $X \to 0^+$, $Y \to \infty$.
   • When $X=1$, $Y = 1/1 = 1$.
   • Therefore, the range of $Y$ is $[1, \infty)$. This immediately eliminates the first and third options.

2. Find the CDF of $Y$, $F_Y(y)$: $F_Y(y) = P(Y \le y)$ Substitute $Y = 1/X$: $F_Y(y) = P(1/X \le y)$ Rearrange the inequality for $X$: $F_Y(y) = P(X \ge 1/y)$

3. Calculate the probability using $X$:

   • We know $X \sim \text{Uniform}[0, 1]$, so its PDF $f_X(x) = 1$ for $0 \le x \le 1$.
   • The probability $P(X \ge 1/y)$ is the integral of $f_X(x)$ from $1/y$ to 1.
   • $P(X \ge 1/y) = \int_{1/y}^{1} f_X(x) ,dx = \int_{1/y}^{1} 1 ,dx$
   • $F_Y(y) = \left[ x \right]_{1/y}^1 = 1 - \frac{1}{y}$

4. Differentiate the CDF to get the PDF: $f_Y(y) = \frac{d}{dy} F_Y(y) = \frac{d}{dy} \left( 1 - \frac{1}{y} \right) = \frac{d}{dy} (1 - y^{-1})$ $f_Y(y) = 0 - (-1)y^{-2} = \frac{1}{y^2}$

5. State the full PDF: $f_Y(y) = \frac{1}{y^2}$, for $1 \le y < \infty$.

Answer: This matches the fourth option.

❓ Question 13

(Using information from the image with questions 12 and 13)

Question: Find the value of $P(Y \le 2)$. Enter the answer correct to one decimal place.

💡 Concept: Calculating Probability

We can solve this in three simple ways.

• Method 1: Use the CDF of $Y$. The definition of the CDF is $F_Y(y) = P(Y \le y)$. From Question 12, we found $F_Y(y) = 1 - \frac{1}{y}$.

• Method 2: Integrate the PDF of $Y$. $P(Y \le 2) = \int_{1}^{2} f_Y(y) ,dy$. We must integrate from 1 (the start of $Y$’s range) up to 2.

• Method 3: Use the original variable $X$. This is often the easiest way.

Solution

Method 1 (Using CDF of Y): $P(Y \le 2) = F_Y(2) = 1 - \frac{1}{2} = \frac{1}{2} = 0.5$

Method 2 (Integrating PDF of Y): $P(Y \le 2) = \int_{1}^{2} \frac{1}{y^2} ,dy = \left[ -\frac{1}{y} \right]_1^2$ $= \left( -\frac{1}{2} \right) - \left( -\frac{1}{1} \right) = -\frac{1}{2} + 1 = \frac{1}{2} = 0.5$

Method 3 (Using X): $P(Y \le 2) = P(1/X \le 2)$ $P(1/X \le 2) = P(X \ge 1/2)$ Since $X$ is uniform on $[0, 1]$, the probability $P(X \ge 1/2)$ is simply the length of the interval $[1/2, 1]$ divided by the total length of the interval $[0, 1]$. $\text{Probability} = \frac{\text{Length}([\frac{1}{2}, 1])}{\text{Length}([0, 1])} = \frac{1 - \frac{1}{2}}{1 - 0} = \frac{\frac{1}{2}}{1} = 0.5$

Answer: 0.5

❓ Question 14

(Using information from the image with questions 14 and 15)

Question: Let the joint Pdf of two random variables $X$ and $Y$ be given by $f_{XY}(x, y) = \begin{cases} 6x^2y, & 0 < x < 1, 0 < y < 1 \ 0, & \text{otherwise} \end{cases}$ Find the marginal of $X$.

💡 Concept: Finding a Marginal PDF

To find the marginal PDF of one variable (e.g., $f_X(x)$), you “integrate out” the other variable ($y$) from the joint PDF. You integrate over the entire range of $y$.

$f_X(x) = \int_{-\infty}^{\infty} f_{XY}(x, y) ,dy$

Solution

1. Set up the Integral: The range of $y$ is from 0 to 1. $f_X(x) = \int_{0}^{1} 6x^2y ,dy$

2. Integrate with respect to $y$: When integrating with respect to $y$, we treat $x$ (and thus $6x^2$) as a constant. $f_X(x) = 6x^2 \left[ \int_{0}^{1} y ,dy \right]$

3. Solve the Integral: $f_X(x) = 6x^2 \left[ \frac{1}{2}y^2 \right]_0^1$ $f_X(x) = 6x^2 \left( \frac{1}{2}(1)^2 - \frac{1}{2}(0)^2 \right)$ $f_X(x) = 6x^2 \left( \frac{1}{2} \right)$ $f_X(x) = 3x^2$

4. State the full PDF: This is valid for $0 < x < 1$. Otherwise, the probability is 0. $f_X(x) = \begin{cases} 3x^2, & 0 < x < 1 \ 0, & \text{otherwise} \end{cases}$

Answer: This matches the fourth option.

❓ Question 15

(Using information from the image with questions 14 and 15)

Question: Find the marginal of $Y$.

💡 Concept: Finding a Marginal PDF

To find the marginal PDF $f_Y(y)$, we “integrate out” the $x$ variable from the joint PDF. We integrate over the entire range of $x$.

$f_Y(y) = \int_{-\infty}^{\infty} f_{XY}(x, y) ,dx$

Solution

1. Set up the Integral: The range of $x$ is from 0 to 1. $f_Y(y) = \int_{0}^{1} 6x^2y ,dx$

2. Integrate with respect to $x$: When integrating with respect to $x$, we treat $y$ (and thus $6y$) as a constant. $f_Y(y) = 6y \left[ \int_{0}^{1} x^2 ,dx \right]$

3. Solve the Integral: $f_Y(y) = 6y \left[ \frac{1}{3}x^3 \right]_0^1$ $f_Y(y) = 6y \left( \frac{1}{3}(1)^3 - \frac{1}{3}(0)^3 \right)$ $f_Y(y) = 6y \left( \frac{1}{3} \right)$ $f_Y(y) = 2y$

4. State the full PDF: This is valid for $0 < y < 1$. Otherwise, the probability is 0. $f_Y(y) = \begin{cases} 2y, & 0 < y < 1 \ 0, & \text{otherwise} \end{cases}$

Answer: This matches the second option.